1. Use lens maker's formula

For a thin lens in air, $P = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ where $\mu$ is the refractive index of the lens material.

2. Power of the given double convex lens

The lens is double convex with equal radii of curvature $R$.

Using sign convention:

• First surface: $R_1 = +R$
• Second surface: $R_2 = -R$

So, $P = (\mu - 1)\left(\frac{1}{R} - \frac{1}{-R}\right)$ $P = (\mu - 1)\left(\frac{1}{R} + \frac{1}{R}\right)$ $P = \frac{2(\mu - 1)}{R}$

3. Power of the plano-convex lens

Let the radius of curvature of the curved surface be $r$.

For a plano-convex lens:

• One surface is plane, so its radius is $\infty$
• Other surface has radius $r$

Hence, $P' = (\mu - 1)\left(\frac{1}{r} - \frac{1}{\infty}\right) = \frac{\mu - 1}{r}$

Given, $P' = 1.5P = \frac{3P}{2}$

So, $\frac{\mu - 1}{r} = \frac{3}{2} \cdot \frac{2(\mu - 1)}{R}$

$\frac{\mu - 1}{r} = \frac{3(\mu - 1)}{R}$

Cancel $(\mu - 1)$: $\frac{1}{r} = \frac{3}{R}$

Therefore, $r = \frac{R}{3}$

4. Check options

The required radius of curvature is $\boxed{\frac{R}{3}}$

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.